Question

Consider a function $f(x)=\int sin(ln \ x)dx$ such that $f(e)=\frac{e}{\sqrt{2}}sin(1-\frac{\pi}{4})$. Then the value of $f(1)$ is

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

Answer: (B)

-\frac{1}{2}

Answer 2

To find the value of $f(1)$, we first need to evaluate the indefinite integral $f(x)=\int \sin(\ln x)dx$.

Step 1: Evaluate the indefinite integral $f(x)=\int \sin(\ln x)dx$.

Let $I = \int \sin(\ln x)dx$. We can use the substitution method. Let $t = \ln x$. Then $x = e^t$, and differentiating both sides with respect to $t$, we get $dx = e^t dt$. Substituting these into the integral: $I = \int \sin(t) e^t dt$.

This is a standard integral of the form $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2}(a \sin(bx) - b \cos(bx))$. In our case, $a=1$ and $b=1$. So, $I = \frac{e^t}{1^2+1^2}(1 \cdot \sin(t) - 1 \cdot \cos(t)) + C'$ $I = \frac{e^t}{2}(\sin(t) - \cos(t)) + C'$.

Now, substitute back $t = \ln x$ and $e^t = x$: $f(x) = \frac{x}{2}(\sin(\ln x) - \cos(\ln x)) + C$.

Step 2: Use the given condition $f(e)=\frac{e}{\sqrt{2}}\sin(1-\frac{\pi}{4})$ to find the constant $C$.

Substitute $x=e$ into the expression for $f(x)$: $f(e) = \frac{e}{2}(\sin(\ln e) - \cos(\ln e)) + C$. Since $\ln e = 1$: $f(e) = \frac{e}{2}(\sin(1) - \cos(1)) + C$.

Now, let's simplify the given value of $f(e)$: $f(e) = \frac{e}{\sqrt{2}}\sin(1-\frac{\pi}{4})$. Using the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $\sin(1-\frac{\pi}{4}) = \sin(1)\cos(\frac{\pi}{4}) - \cos(1)\sin(\frac{\pi}{4})$. We know that $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. So, $\sin(1-\frac{\pi}{4}) = \sin(1)\frac{1}{\sqrt{2}} - \cos(1)\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(\sin(1) - \cos(1))$.

Substitute this back into the expression for $f(e)$: $f(e) = \frac{e}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(\sin(1) - \cos(1))$ $f(e) = \frac{e}{2}(\sin(1) - \cos(1))$.

Now, equate the two expressions for $f(e)$: $\frac{e}{2}(\sin(1) - \cos(1)) + C = \frac{e}{2}(\sin(1) - \cos(1))$. From this equation, it is clear that $C=0$.

So, the definite form of the function is: $f(x) = \frac{x}{2}(\sin(\ln x) - \cos(\ln x))$.

Step 3: Calculate the value of $f(1)$.

Substitute $x=1$ into the function $f(x)$: $f(1) = \frac{1}{2}(\sin(\ln 1) - \cos(\ln 1))$. We know that $\ln 1 = 0$. So, $f(1) = \frac{1}{2}(\sin(0) - \cos(0))$. We know that $\sin(0) = 0$ and $\cos(0) = 1$. $f(1) = \frac{1}{2}(0 - 1)$ $f(1) = -\frac{1}{2}$.

The final answer is $-\frac{1}{2}$.

Explanation of the solution:

1. Integrate $f(x)$: The integral $\int \sin(\ln x) dx$ is solved by substitution $t = \ln x$, transforming it into $\int e^t \sin t dt$. This is a standard integral form $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2}(a \sin(bx) - b \cos(bx))$. Applying this formula with $a=1, b=1$ and then substituting back $t=\ln x$ yields $f(x) = \frac{x}{2}(\sin(\ln x) - \cos(\ln x)) + C$.

2. Determine Constant $C$: Use the given condition $f(e)=\frac{e}{\sqrt{2}}\sin(1-\frac{\pi}{4})$. Substitute $x=e$ into $f(x)$ to get $f(e) = \frac{e}{2}(\sin(1) - \cos(1)) + C$. Simplify the given $f(e)$ using the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, which gives $\frac{e}{\sqrt{2}}\sin(1-\frac{\pi}{4}) = \frac{e}{2}(\sin(1) - \cos(1))$. Equating the two expressions for $f(e)$ reveals $C=0$.

3. Calculate $f(1)$: With $C=0$, $f(x) = \frac{x}{2}(\sin(\ln x) - \cos(\ln x))$. Substitute $x=1$ into this function. Since $\ln 1 = 0$, $f(1) = \frac{1}{2}(\sin(0) - \cos(0))$. Using $\sin(0)=0$ and $\cos(0)=1$, we get $f(1) = \frac{1}{2}(0 - 1) = -\frac{1}{2}$.